Question
Solve:
$\frac{2-7\text{x}}{1-5\text{x}}=\frac{3+7\text{x}}{4+5\text{x}}$
$\frac{2-7\text{x}}{1-5\text{x}}=\frac{3+7\text{x}}{4+5\text{x}}$
✓
Answer
$\frac{2-7\text{x}}{1-5\text{x}}=\frac{3+7\text{x}}{4+5\text{x}}$
$\Rightarrow (4+5\text{x})(2-7\text{x}) = (1-5\text{x})(3+7\text{x})$
$\Rightarrow 8 - 28\text{x} + 10\text{x} - 35\text{x}^2 = 3 + 7\text{x} - 15\text{x} - 35\text{x}^2$
$\Rightarrow-35\text{x}^2 - 18\text{x} +8 = -35\text{x}^2 - 8\text{x} + 3$
$\Rightarrow-35\text{x}^2 + 35\text{x}^2 -18\text{x} + 8\text{x} = -8 + 3$
$\Rightarrow-10\text{x} = -5$
$\Rightarrow \text{x} = -5-10= \frac{1}{2} $
$\therefore \text{x} = \frac{1}{2}$
$\Rightarrow (4+5\text{x})(2-7\text{x}) = (1-5\text{x})(3+7\text{x})$
$\Rightarrow 8 - 28\text{x} + 10\text{x} - 35\text{x}^2 = 3 + 7\text{x} - 15\text{x} - 35\text{x}^2$
$\Rightarrow-35\text{x}^2 - 18\text{x} +8 = -35\text{x}^2 - 8\text{x} + 3$
$\Rightarrow-35\text{x}^2 + 35\text{x}^2 -18\text{x} + 8\text{x} = -8 + 3$
$\Rightarrow-10\text{x} = -5$
$\Rightarrow \text{x} = -5-10= \frac{1}{2} $
$\therefore \text{x} = \frac{1}{2}$
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